Is there an isometry of $\Bbb R^2$ taking the curve given by $\alpha(t) = (1+\cos(t), 2+\sin(t))$ for $t \in [0, \pi]$ to the curve given by $\beta = (t, \sin(t))$ for $t \in [0, c]$ where $c$ is some constant?
In order to prove there's an isometry, must I prove that there exists a $2 \text{x} 2$ matrix $A$ so that $\beta(t) = A * \alpha(t)$. If I prove such a matrix doesn't exist, is that sufficient to prove that there isn't the existence of an isometry, or is there a different way to prove/disprove the existence of an isometry?
The very first thing I would do is sketch the two curves in the plane.
If by "isometry of $\mathbb R^2$" we mean an isometry under the usual Euclidean metric, then it suffices to show that the distances between the endpoints of each curve is not preserved except for a specific value of $c$; and then, in this case, show that the distance between an endpoint and the respective midpoint is not preserved.